Question

Equation of plane through the line of intersection of the planes $2x+3y-4z=1$ and $3x-y+z+2=0$ and it makes an intercept of $4units$ on the positive $x$-axis is $2x+3y-4z-1+\lambda (3x-y+z+2)=0$, then $\lambda$ is

• A. $\frac{1}{2}$
• B. $30$
• C. $7$
• D. $-\frac{1}{2}$

Answer 1

The correct option is D $-\frac{1}{2}$

The $x-$intercept can be found by substituting $y=0$ and $z=0$.

Hence, we get $(2+3\lambda )4=1-2\lambda$.

$\Rightarrow 1-2\lambda =8+12\lambda$

$\Rightarrow -14\lambda =7\Rightarrow \lambda =\frac{-1}{2}$